Statement I: For f : { 1 , 2 , 3 , 4 , 5 } → { 1 , 2 , 3 , 4 , 5 } , the number of onto functions satisfying f ( i )  = i for every i is 42 . Statement II: The number of derangements of n objects is n ! ∑ k = 0 n ( − 1 ) k / k ! .

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MathematicsPermutation & CombinationDearrangementMedium2 minQB

MathematicsMediumsingle choice

Statement I: For $f : {1, 2, 3, 4, 5} \to {1, 2, 3, 4, 5}$ , the number of onto functions satisfying $f (i) \neq = i$ for every $i$ is 42 . Statement II: The number of derangements of $n$ objects is $n! \sum_{k = 0}^{n} (- 1)^{k} / k!$ .

Options:

Answer:

Solution:

Onto self-maps of a five-element set are permutations. The derangement formula in II gives $! 5 = 44$ , not 42, so I is false.

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:Dearrangement

⏱ 2mℹ️ Source: QB

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